Question

A woman can read the large print in a newspaper only when it is at a distance of 58 cm or more from her eyes. (a) Is she nearsighted (myopic) or farsighted (hyperopic), and what kind of lens is used in her glasses to correct her eyesight? (b) What should be the refractive power (in diopters) of her glasses (worn 2.3 cm from the eyes), so she can read the newspaper at a distance of 29 cm from the eyes?

Answer 1

A) Without a given age, we can assume she is hyperopic since the near point is 25 cm from the retina in someone in their early 20s. Since her near point is further than that, she is unable to see objects closer than that clearly, and is farsighted.

B)

Because the eyeglasses are a distance away from her eyes, you have to factor them into both the image and object distance in the equation 1/f (focal point) = 1/di (image distance) + 1/do (object distance). The reason each of them are factored in is because the equation accounts for the lens of the glasses, not that of the eye. This causes the object distance (where she wants to read the paper at, 29 cm) to be 2.3 cm closer to the glasses' lens than her eyes, and her near point (the distance she can read the paper at without glasses) to similarly be 2.3 cm closer to the glasses' lens than her eyes.

In this case:
1/f = 1/di + 1/do
1/f = 1/-(58-2.3) + 1/(29-2.3)
1/f= 1/-(55.7) + 1/26.7

f = 51.3 cm

Note that the distance of the object is negative because it's to the left of the lens of the eyeglass. Now to get the answer in diopters, you have to use the equation R(refractive power) = 1/f (in meters), which is 1/0.513, or 1.949 diopters